Inverse Laplace transform — partial fractions, tables

4.6.29 · Maths › Ordinary Differential Equations

WHAT hai inverse Laplace transform?

WHY yeh well-defined hai? Un functions ke liye jo piecewise continuous hain aur exponential order ki hain, forward transform one-to-one hota hai (Lerch's theorem), isliye "undo" karne par ek unique $f (t)$ milta hai. Yahi uniqueness hai jo hume confidence ke saath ek lookup table use karne deti hai.

Core table (derive karo, andhe hokar mat yaad karo)

Har entry forward integral $F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$ se aati hai. Main key ones derive karta hoon taaki table earned lage.

Master table (yeh 7 yaad karo):

$F (s)$	$f (t) = L^{- 1} {F}$
$\frac{1}{s}$	$1$
$\frac{1}{s - a}$	$e^{a t}$
$\frac{1}{s ^{n + 1}}$	$\frac{t ^{n}}{n !}$
$\frac{ω}{s ^{2} + ω ^{2}}$	$sin ω t$
$\frac{s}{s ^{2} + ω ^{2}}$	$cos ω t$
$\frac{ω}{( s - a ) ^{2} + ω ^{2}}$	$e^{a t} sin ω t$
$\frac{s - a}{( s - a ) ^{2} + ω ^{2}}$	$e^{a t} cos ω t$

HOW: partial fractions — recognition machine

Ek proper rational $F (s) = \frac{P ( s )}{Q ( s )}$ (jahan degree of $P$ < degree of $Q$ ) ko $Q (s)$ ke factors ke basis par split kiya jaata hai:

Distinct linear $(s - a)$ → term $\frac{A}{s - a}$
Repeated linear $(s - a)^{k}$ → terms $\frac{A _{1}}{s - a} + \frac{A _{2}}{( s - a ) ^{2}} + \dots + \frac{A _{k}}{( s - a ) ^{k}}$
Irreducible quadratic $(s^{2} + b s + c)$ → term $\frac{B s + C}{s ^{2} + b s + c}$ (phir shift table se match karne ke liye complete the square)

Cover-up method (simple poles ke liye fast coefficient finder): factor $(s - a)$ ke liye $A$ paane ke liye, dono sides ko $(s - a)$ se multiply karo aur $s = a$ rakh do: $A = [(s - a) F (s)]_{s = a} .$ WHY kaam karta hai: multiply karne se apne term ke denominator mein $(s - a)$ khatam ho jaata hai, aur $s = a$ par har doosre term mein abhi bhi $(s - a)$ factor hota hai → woh vanish ho jaate hain, aur $A$ isolate ho jaata hai.

Worked Example 1 — distinct linear factors

Find $L^{- 1} {\frac{3 s + 1}{( s - 1 ) ( s + 2 )}}$ .

Step 1. Likho $\frac{3 s + 1}{( s - 1 ) ( s + 2 )} = \frac{A}{s - 1} + \frac{B}{s + 2}$ . Yeh step kyun? Do distinct linear poles → do simple-pole atoms.

Step 2. Cover-up: $A = \frac{3 ( 1 ) + 1}{1 + 2} = \frac{4}{3}$ , $B = \frac{3 ( - 2 ) + 1}{- 2 - 1} = \frac{- 5}{- 3} = \frac{5}{3}$ . Yeh step kyun? Cover-up har residue ko bina system solve kiye isolate karta hai.

Step 3. Har atom ko $L^{- 1} {1/ (s - a)} = e^{a t}$ se invert karo: $f (t) = \frac{4}{3} e^{t} + \frac{5}{3} e^{- 2 t} .$ Yeh step kyun? Linearity hume term-by-term invert karne deti hai.

Worked Example 2 — irreducible quadratic (complete the square)

Find $L^{- 1} {\frac{s + 3}{s ^{2} + 2 s + 5}}$ .

Step 1. Complete the square: $s^{2} + 2 s + 5 = (s + 1)^{2} + 4$ . Yeh step kyun? Denominator ke real roots nahi hain; shift table ko form $(s - a)^{2} + ω^{2}$ chahiye jahan $a = - 1, ω = 2$ .

Step 2. Numerator ko $(s + 1)$ ke around rewrite karo: $\frac{s + 3}{( s + 1 ) ^{2} + 4} = \frac{( s + 1 ) + 2}{( s + 1 ) ^{2} + 4} = \frac{s + 1}{( s + 1 ) ^{2} + 4} + \frac{2}{( s + 1 ) ^{2} + 4} .$ Yeh step kyun? $\frac{s - a}{( s - a ) ^{2} + ω ^{2}}$ ( $e^{a t} cos$ deta hai) aur $\frac{ω}{( s - a ) ^{2} + ω ^{2}}$ ( $e^{a t} sin$ deta hai) se match karo. Doosre term mein $2 = ω$ hai, perfect.

Step 3. Table padhao $a = - 1, ω = 2$ ke saath: $f (t) = e^{- t} cos 2 t + e^{- t} sin 2 t .$ Yeh step kyun? First-shift theorem shifted denominator ko $e^{- t}$ envelope mein convert karta hai.

Worked Example 3 — repeated factor

Find $L^{- 1} {\frac{s}{( s + 1 ) ^{2}}}$ .

Step 1. $\frac{s}{( s + 1 ) ^{2}} = \frac{A}{s + 1} + \frac{B}{( s + 1 ) ^{2}}$ . Yeh step kyun? Double pole ko dono powers chahiye; sirf ek term se underfit ho jaata.

Step 2. $(s + 1)^{2}$ se multiply karo: $s = A (s + 1) + B$ . $s = - 1$ rakh do: $- 1 = B$ . $s$ -coefficient compare karo: $1 = A$ . Yeh step kyun? Cover-up highest-power coefficient directly deta hai; matching baaki deta hai.

Step 3. Invert karo: $L^{- 1} {1/ (s + 1)} = e^{- t}$ , aur $L^{- 1} {1/ (s + 1)^{2}} = t e^{- t}$ ( $1/ s^{2} \to t$ ka shift): $f (t) = e^{- t} - t e^{- t} = (1 - t) e^{- t} .$

Common mistake Steel-manned common mistakes

(1) Repeated-power term bhool jaana. Yeh lagta hai sahi hai ki sirf $\frac{s}{( s + 1 ) ^{2}} = \frac{A}{( s + 1 ) ^{2}}$ likho — "ek factor, ek term." Kyun tempting lagta hai: distinct factors ne exactly ek-ek term diya tha. Fix: factor $(s - a)^{k}$ se $k$ terms aate hain, ek per power $1 \dots k$ , kyunki numerator degree $< k$ ka koi bhi polynomial ho sakta hai. (2) Improper fraction ko directly invert karne ki koshish. Agar $de g P \geq de g Q$ hai, toh table apply nahi hota. Kyun tempting lagta hai: aap seedha partial fractions par jaana chahte ho. Fix: pehle polynomial long division karo; polynomial part $δ (t)$ -type/derivative terms mein invert hota hai, remainder proper hota hai. (3) Sine numerator galat match karna. $L^{- 1} {s / (s^{2} + 9)} = sin 3 t$ likhna. Kyun tempting lagta hai: $s^{2} + 9$ "sine jaisa lagta hai." Fix: upar $s$ → $cos$ ; upar constant $ω = 3$ → $sin$ . Hamesha check karo ki numerator $s$ carry karta hai ya $ω$ . (4) Completing the square ke baad galat $ω$ . $(s + 1)^{2} + 4$ mein $ω = 4$ use karna. Fix: table $ω^{2}$ use karta hai, toh $ω = 4 = 2$ , aur sine numerator wahi $ω$ hona chahiye.

Recall Feynman: ek 12-saal ke bachche ko samjhao (hidden)

Socho tumne ek English sentence ko ek secret code mein translate kiya taaki use aasaani se rearrange kar sako. Ab tumhe use wapas English mein translate karna hai. Tum ek saath bade chunks translate nahi karte — tum har code-word apni dictionary mein dhundhte ho. Partial fractions woh scissors hai jo lambe coded message ko single code-words mein kaat deti hai, aur table tumhari dictionary hai. Pieces mein kaat lo, har piece lookup karo, English wapas jod lo. Ho gaya.

Forecast-then-Verify drill

Table check karne se pehle, predict karo $L^{- 1} {\frac{1}{s ^{2} - 4}}$ ki form kya hogi. $\pm 2$ par real poles? → $e^{\pm 2 t}$ expect karo, matlab $sinh / cosh$ , $sin$ nahi. Phir verify karo: $\frac{1}{s ^{2} - 4} = \frac{1}{4} (\frac{1}{s - 2} - \frac{1}{s + 2}) \Rightarrow \frac{1}{4} (e^{2 t} - e^{- 2 t}) = \frac{1}{2} sinh 2 t$ . ✓ Andar ke constant ka sign ( $- 4$ vs $+ 4$ ) trig↔hyperbolic flip kar deta hai.

Flashcards

1/ (s - a)

ka inverse Laplace transform kya hai?

e^{a t}

L^{- 1} {s / (s^{2} + ω^{2})}

kya hai?

cos ω t

L^{- 1} {ω / (s^{2} + ω^{2})}

kya hai?

sin ω t

L^{- 1} {1/ s^{n + 1}}

kya hai?

t^{n} / n!

Inverses ke liye first shift theorem batao.

L^{- 1} {F (s - a)} = e^{a t} f (t)

jahan

f = L^{- 1} {F}

Factor

(s - a)^{k}

kitne partial-fraction terms contribute karta hai?

k

terms, powers

1

k

tak

Partial fractions se pehle kya karna chahiye agar

de g P \geq de g Q

ho?

Polynomial long division (pehle proper banao)

Factor

(s - a)

ke liye cover-up coefficient?

A = [(s - a) F (s)]_{s = a}

L^{- 1} {1/ ((s - a)^{2} + ω^{2}) \cdot ω}

e^{a t} sin ω t

L^{- 1} {1/ (s^{2} - a^{2})}

\frac{1}{a} sinh a t

Numerator mein

s

vs constant

ω

: kaun cos deta hai kaun sin?

s

-on-top →

cos

;

ω

-on-top →

sin

L^{- 1}

unique kyun hai?

Lerch's theorem — forward transform piecewise-continuous functions of exponential order ke liye one-to-one hota hai

Connections

Laplace Transform — definition and existence
Solving ODEs with Laplace Transforms
Partial Fraction Decomposition (Algebra)
First and Second Shift Theorems
Convolution Theorem
Heaviside Step & Dirac Delta
Hyperbolic functions sinh/cosh vs sin/cos